Differentiation: Applications of derivatives
Increasing and decreasing
Increasing and decreasing
A function#f# is #\blue{\text{increasing}}# in #x# if #f(x)# increases as #x# increases.
A function #f# is #\green{\text{decreasing}}\ (dashed)# in #x# if #f(x)# decreases as #x# increases.
In the example, we see that a function also both increase and decrease. We say that the function increases on the interval #\ivoo{-\infty}{6}# and decreases on the interval #\ivoo{6}{\infty}#.
We can check whether a function increases or decreases in a point #x# by looking at the derivative in that point.
A function #f# #\blue{\text{increases}}# in a point #x# if #f'(x)\gt 0#.
A function #f# #\green{\text{decreases}}# in a point #x# if #f'(x)\lt 0#.
A function #f# can transition from #\blue{\text{increasing}}# to #\green{\text{decreasing}}# (and the other way around) in a point #p# if #f'(p)=0#.
Example
\begin{array}{rcll}f(x)&=&x^3-8x& \text{with } \quad f'(x)=3x^2-8\\
f'(2)&=&4&\quad\text{therefore }f\blue{\text{ increases }}\text{in }x=2\\f'(1)&=&-5&\quad\text{therefore }f\green{\text{ decreases }}\text{in }x=1\end{array}
Step-by-step | Example | |
We want to determine the interval or intervals at which the function #f# #\blue{\text{increases}}#. |
#f(x)=\left(x-4\right)^2+6# | |
Step 1 |
Determine the derivative of #f#. |
#f'(x)=2\left(x-4\right)# |
Step 2 |
Determine the zeros of the derivative. |
#x=4# |
Step 3 |
For points to the left and right of the zeros, determine whether #f'# is positive or negative. |
#f'(0)=-8# and #f'(6)=4# |
Step 4 |
Now determine the interval / intervals on which #f# increases. The function #f# increases if #f'(x) \gt 0#. |
#f# #\blue{\text{increases}}# on #\ivoo{4}{\infty}# |
Step 1 | We determine the derivative of #f# using the power rule. This gives: \[f'(x)=2x-9\] |
Step 2 | We solve the equation \[2x-9=0\] This goes as follows: \[\begin{array}{rcl}2x-9&=&0 \\ &&\phantom{xxx}\blue{\text{the equation we need to solve}} \\ 2x&=&9 \\ &&\phantom{xxx}\blue{\text{both sides minus }-9} \\ x&=&\displaystyle {{9}\over{2}} \\ &&\phantom{xxx}\blue{\text{both sides divided by }2} \\\end{array}\] |
Step 3 | #f'(2)=-5# # f'(6)=3# |
Step 4 | Therefore, the function #f# is decreasing on the interval #\ivoo{-\infty}{{{9}\over{2}}}# and increasing on the interval #\ivoo{{{9}\over{2}}}{\infty}#. Hence, #a={{9}\over{2}}#. |
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